This quantity [itex]c[/itex] is clearly another integral of motion. But- in general - this does not isolate the region where the motion takes place any further, because [itex]\cos^{-1}z[/itex] is a multiple-valued function. To see this more clearly, let us write

Where [itex]Cos^{-1}z[/itex] (with an uppercase C) denotes the principal value. For a given value of [itex]y[/itex] we will get an infinite number of [itex]x[/itex]'s as we take [itex]n=0, \pm 1, \pm 2, \dots [/itex]

Thus, in general, the curve will fill a region in the [itex](x,y)[/itex] plane.

A special situation arises if [itex](\omega_{x}/\omega_{y})[/itex] is a rational number. In that case, the curve closes on itself after a finite number of cycles. Then [itex]c[/itex] is also an isolating integral and we have three isolating integrals: [itex](E_{x}, E_{y}, c)[/itex]. The motion is confined to closed (one-dimensional) curve on the surface of the torus.

Click to expand...

This last part is not still clear to me.

Can someone please explain why a rational ratio of frequencies make a candidate integral of motion single valued and therefore the motion takes place on a closed (one dimensional) curve on the surface of the two torus?

My confusion was that in the case of a rational ratio, even though periodic, we have still multiple (finitely) values and not a single valued variable.

Turns out that like the nth root of unity in complex plane we can define a single valued function over a multiple (finitely) valued variable. And therefore in the case of a rational ratio we have a new isolating integral of motion which limits the dimension of phase space to just one instead of four.

Clearly in the case of the irrational ration you cannot have a periodic valued variable and therefore the phase space dynamic covers the whole surface of the two torus.